Now consider how those same equations for discharging a capacitor apply to charging the capacitor. Also realize that this discussion only looks at instantaneous time. When looking over some duration of time, the value of V and I are changing so that one must resort to using calculus.
The resistance R is discharging the capacitor C, so the voltage V across the capacitor is decreasing to zero. Therefore dV/dt must have a negative value.
The equation is also related to the definition of capacitance:
C=Q/V = dQ/dV = Idt/dV or C(dV/dt)=I
Check out Maxim application note 742, "Impedance Matching and the Smith Chart: The Fundamentals"
http://www.maxim-ic.com/app-notes/index.mvp/id/742
http://rfdesign.com/images/archive/0700Chan52.pdf
I don't see any way to get a clean transfer function due to the nonlinearity of the MOSFET characteristic function. However, if you assume that the constant terms in the transfer function are not important, e.g. use AC coupling, and that the non-linear parts of the transfer function are...
Reading the datasheet suggests these RAM chips were intended to be connected in parallel for expansion, ie, A0->A0....A12->A12, I/On->I/On, CE1->CE1, WE->WE, OE->OE. But the CE2 line will multiplex the RAM chips. See if you can make that work.
What do you mean by 'digital' signal? A 5V CMOS digital signal is somewhat different than a 5V TTL digital signal. Is it the 12 V that is switching at 1.6 Hz or do you need a circuit to generate the 1.6 Hz? Maybe all you need is a 5V regulator and 555 timer.
Three equations govern this source follower:
Vin=VGS+Vo
Vo=Ids x Rs
Ids=(K/2)(VGS-VT)^2
Solving for Vo yields a quadratic solution, see attachment. It would seem that only one of the solutions should correspond to a condition that VGS>VT.
Also the source follower output has an...
The problem started with an equation giving I as a function of V, I=f(V). So if you take the derivative dV/dI you get another function of V giving the incremental resistance for any value of V.
A numerical approximation needs to provide the value dV/dI for any value of V using the same...
To find the bias point for this circuit requires two equations. There is the characteristic equation for the MOSFET relating VGS to drain current, and the bias equation relating bias voltage to drain current.
ID=(K/2)(VGS-VT)^2
VI=VGS+IDxRs
Note that the characteristic equation is...
Initially I used my HP-15C calculator. But then I just entered the formula into the Open Office Calc spreadsheet for a sensitivity analysis of how much the final answer changed using 4.01, 4.001, 4.0001 volts.
Have a look at Cypress AN-2336 "Simplified FSK Detection" which describes demodulation using signal correlation. Seems pretty interesting, and is rather simple to implement.
N2336 - PSoC® 1 - Simplified FSK Detection http://www.cypress.com/?rID=2735
Is that a guess, or did you fail to read the hint provided by Harald?
VI+Vi-Vgs-Id*Rs = 0V or Vgs = (VI+Vi)-Id*Rs
Vgs is defined as the voltage from the gate to the source, Vgs=Vg-Vs, where in this case, Vg = (VI+Vi) and Vs is the voltage across the source resistance so Vs= Id*Rs...
When you are asked to find the bias point, the input signal voltage vi is quiesced to zero. The bias point is also referred to as the quiescent point, hence vi=0.
A simple method for doing it the hard way is to write the node equations (in this case write one equation since there is only one node in the circuit) then have a symbolic algebra engine create the Bode plot. One can examine the transfer function of the circuit to determine a frequency where...
@Harald:
Closed form solutions are truly more elegant. Yet it has been so many years since I took the derivative of anything, I would need to open my old calculus textbook. Still understand the principles behind derivatives but have forgotten the mechanics of doing it. Also, having a quick...
Not the answer I got, but I did not try to get fancy by finding the derivative. Just used numerical approximation instead with the formula I=0.002 x V^(3/2). Then found the current at 4 volts and at 4.001 volts so delta:V=0.001 and delta:I is as calculated. Same process at 10 volts.
If by 'extremely exact' you mean an ideal voltmeter, then the meter would have infinite resistance. Voltmeter 1 would read 6 V while Voltmeter 2 would read zero.